PROGRESS TEST-5 · 2017-12-02 · [ 8 ] PT-V (ADV) RBA_11.11.2017 Mentors Eduserv: Parus Lok...

PROGRESS TEST-5RBA

(JEE ADVANCED PATTERN)

Test Date: 11-11-2017

[ 2 ] PT-V (ADV) RBA_11.11.2017

Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna-1Helpline No. : 9569668800 | 7544015993/4/6/7

CHEMISTRY1. (B)

PV = nRT .................(1)P = 3170 Pa, V = 1 dm3 = 10–3 m3,R = 8.314 JK–1 mol–1, T = 300 K

From eqn. ...............(1)33170 10 n 8.314 300

3n 1.27 10 mol

2. (D)3. (B)4. (C)

23 = 8.5. (B)

2 3 3 3 3I II IVIII

K CO CaCO MgCO BeCO

6. (D)

0 (exception)

7. (C)If 2s – 2p mixing is not considered for Li2 to N2

z x y

x y

2 *2 2 *2 2 1 11s 1s 2s 2s 2p 2p 2p

* *2p 2p z2p

C2 total e– = 12

PT-V (ADV) RBA_11.11.2017 [ 3 ]


8. (A)

cycle cycle cycleq w 0 w 5 J

AB BC CAw w w 5

CA10(2 1) 0 w 5

CAw 5 J

9. (A, B, C)Cooling effect is observed when the real gas undergoes adiabatic expansion below theinversion tempeature.

10. (B), (C)11. (A, C, D)

Equivalent weight = Molecular weight

n factor

(A) n–factor of 2 3H PO as acid = 1

as base = 1

In both case, eq. wt. of 2 381H PO 811

(B) n–factor of 2 4H PO as acid = 1 or 2

as base = 1(C) In acidic medium

Eq. wt. of KMnO4 = Molecular weight

5In basic medium

Eq. wt. of 4Molecular weightKMnO

1

In neutral/slightly basic medium

Eq. wt. of 4Molecular weightKMnO

3

(D) In 2MgH , O.S. of H 1

In 2 2H O , O.S. of H 1

12. (A,B,C)(A) When (Na) sedium react with excess ammonia give a blue colour paramagnetic solution.

[ 4 ] PT-V (ADV) RBA_11.11.2017


(B) 2 2

2

K O KO

K

2O

Paramagnetic(C) 3Cu + 8HNO33Cu(NO3)2 + 2NO + 4H2O (Dilute)Where NO is paramagnetic

(D)CH +CH2 3

+O (air) 2

CH +CH2 3+H O2 2

(2 - Ethyle anthraquinol)Where H2O2 is Diamagnetic

13. (9)

(1) (2) (3) (4)

(5) (6) CH3–CH = C = CH2 (7) CH2 = CH – CH = CH2

(8) CH3CC–CH3 (9) CHC–CH2–CH3

14. (4)

(1) CH3—CH2—CH2—CH2—CH2—OH (2) OH—CH—CH—CH—CH|CH

223

3

(3) OH—CH—CH—CH—CH|CH

223

3

(4)

3

23

3

CH|

OH—CH—C—CH|CH

15. (2)

2 6 3

2 4 2 22 4

2 3 3 2

2 4 2 2

Cr(H O) Cl

Cr(H O) Cl Cl.2H Oare not affected by conc.H SO

[Cr(H O) Cl ].3H O[Cr(H O) Cl ].2H O

PT-V (ADV) RBA_11.11.2017 [ 5 ]


16. (3)

H nC T

38 8.31 30 3000 J 3 kJ2

17. (4)

2 x 2 2

0.220g 0.045g

M O (Metallic oxide) H M(s) H O( )

POAC on O

0.220 0.045x 1

2M 16x 18

3.96x 0.09 M 0.72x

ME 36x

E 49

18. (5)

2 2NH NH

is a monodontate ligands due to formation of unstable chelation and all other com-pounds are chelating ligands.

19. (A - q, s) ; (B - r) ; (C - p, t) ; (D - t)20. (A - r), (B - s) ; (C - p) ; (D - q)

PT-V (ADV) RBA_11.11.2017[ 6 ]

Mentors Eduserv: Parus Lok Complex, Boring Road Crossing, Patna-1 Helpline No. : 9569668800 | 7544015993/4/6/7

PHYSICS 21. Applying Snell’s law between the points O and P, we have

200

1290sin60sin2H

, 2121

232

H

3

21 2 H ,

1

32H

(A)

22. Apparent position of the object w.r.t lens.

cm152

101

10

u

Rv

15.121151

v = 7.5 cm

(A)

23. (B)

Take the mass m as a point mass. At the instant when the pendulum collides with the nail, m has

a velocity v = 2g . The angular momentum of the mass with respect to the point at which the nail

locates is conserved during the collision. Then the velocity of the mass is still v at the instant after

the collision and the motion thereafter is such that the mass is constrained to rotate around the

nail. Under the critical condition that the mass can just swing completely round in a circle, the

gravitational force when the mass is at the top of the circle. Let the velocity of the mass at this

instant be v1, and we have

d

mv21

= mg,

or v12 = ( – d)g

The energy equation

2

mv 2 = )d(mg2

2mv2

1 ,

or 2g = ( – d)g + 4( – d)g

then gives the minimum distance as

d = 53

PT-V (ADV) RBA_11.11.2017 [ 7 ]


24. (A) The plumb-line is so set up that the resultant of its weight mg and the tension in the thread T

produces a centripetal force F = m2R (fig.). Clearly R = r + sin . Therefore,

2 =

sinr

tang

, =

sinr

tang

.

25. As F1 – F2 < 2Mg, so system will not accelerate. Again here F1 > F2, so block A is the driving

block and block B is driven block. So friction on block A acts towards left but in the block B it may

act left or right.

(B)

26. (A)

27. The equivalent capacitance

)4(4)34(

)43(40

0 aaa

aaC

a064

(A)

28. Ceq = 3/2 F

Charge flow q = Ceq (10 – 3

15 ) = 23 × 5 = 7.5 C

(B)

29. (C,D)

30. (B,C)

When current is maximum then it is possible when this is the circuit.

R

F

T

r

mg

24V1

PT-V (ADV) RBA_11.11.2017[ 8 ]


i = 24 amp

Now, no current flow through AB.

Therefore potential difference between A and B is zero.

Current is minimum only this is the circuit

Req = 6 + 1 + 4.5 + 0.5 = 12

i = 1224 = 2 Amp

VA – VB = 1 × 1 = 1 volt

31. Area 2

461021 2v

v = 10 m/s

Area upto 30 m 2

63021 2v

1802 v

maxv 180 14

(B) and (C)

32. When t = 3s block just about to move and acceleration

of block given by 1

2

ta t > 3

dttdvv

210

30

6

2920502

2

10

3

2

ttv

5.315.130 m/s

(A) (B) and (C)

A

B

6

32Amp 2A

mp

6

1 1

1

9

6

30

a(m/s)2

S(m)

4

10

B A F=t

fk= 1N fs= 1N

fk= 1N fs= 2N

PT-V (ADV) RBA_11.11.2017 [ 9 ]


33. (5)

If C doesn’t move then aA = 4aB … (i)

P – T – µmg = m 4aB …(ii)

4T – mg = maB …(iii)

mgp = 5

34. (8)

dr.FU

r

kmU

Ki +Ui = Kf +Uf

2/R3

Km0

2R

Kmmv21 2

R3Km2

RKm2

2mv2

R3

Km42

mv2

R3K8V ,V = 8 m/s

35. (6)

FE = qE = 11 N

Fg = mg = 5N

So Net force = F = 6N upward

geff = 5.0

6mF 12 m/s2

so Vmin = effg5 = )1060(125 2

so Vmin = 6 m/sec

36. (2)

|| =

20 )/(1

11lR

q . The sign of depends on how the direction of the normal to the circle

is chosen.

uR/2

PT-V (ADV) RBA_11.11.2017[ 10 ]


37. (4)

qVa = qVb + 21 mv2

2.0 × 10–9 × 9 × 109

2103

1103 99

× 100

= 2.0 × 10–9 × 9 × 109

2103

1103 99

×100 + 21 × 5.0 × 10–9 v2

10–9 × 18

23 × 100 = 18 × 10–9 × 100

23

+

21 × 5.0 × 10–9v2

1800

23

23 =

21 × 5.0 × v2

5

61800 = v2

360 × 6 = v2 6 × 6 × 10 × 6 = v2

12 15 = v

38. (3)

uvuv

u

6116

uv

uvuvu

66

6 2v 6u v u

uvuuv 626 64266 vuuv

u = 3 km/h

39. A p ; B r ; C q ; D s

40. A q ; B p ; C r ; D q

In parallel current distributes in inverse ratio of resistor. Hence, distribution of current in different

resistors is as shown in figure.

For power generation apply P = i2R.

3Ri =

RPD =

410 = 2.5 A

4Ri =

RPD =

25 = 2.5 A

i31

2

4

5

3

i32

i

i65

i83

PT-V (ADV) RBA_11.11.2017 [ 11 ]


MATHEMATICS41. (D)

Let log3n = x

y = 5x2 – 12x + 9

y is minimum at x = – a2b

= 1012

= 56

Here log3n = 56

n = 36/5 3.70

which is not natural hence minimum occurs at the closest integer

now 4 > 36/5

45 > 36

1024 > 729 which is true

42. (B)

2

h 0 h 0

f x h f x f h | x |h x hf ' x lim lim

h h

Also x = y = 0 f 0 0

h 0

f h f 0f ' x lim | x | xh

h

f ' x f ' 0 | x |

43. (A)

f x 1 x 1 x 1 x 3 1 x x 2 x 1 f ' x 1

44. (A)2 2 2 4sin xcos x cos xsin x 1

2 2 2sin xcos x 1 sin x 1

2 4sin xcos x 1 , No value of ‘x’

[ 12 ] PT-V (ADV) RBA_11.11.2017


45. (C)

For cx Q , f( ) 0 and xLimf(x)

is also zero because if x moves towards and attaining

irrational value then it is zero, or if attaining rational value then the denominator can be madeas large as possible.

46. (C)

g(x) x 30 x –3g(x) –3 x –2

f(x) 0 x 2g(x) 2 x 1

0 x 1g(x) x 1

47. (D)The image of A in y = x will lie on BCA’ = (5, 4)

AD BC

A (4,5)

E

B (h, k) D A'C

(3,3)

4 k2 15 h

8 2k 5 h

h k

13h k3

48. (D)

Let P (acos ,asin )

and centroid of APB be (h, k).

Then acos 0 a asin a 0h ,k3 3

P

y

A (0,a)

B (a,0) x

3h 3kcos 1, sin 1a a

2 2sin cos 1

PT-V (ADV) RBA_11.11.2017 [ 13 ]


2 23h 3k1 1 1a a

2 2 29h 9k 6ah 6ak a 0

so locus of centroid is

2 2 29x 9y 6ax 6ay a 0

49. (A, D)Let us take distance between paralle sides then BM + AL = 5

A B

CD

L M

5

5

2 5

10

2 225 20 5 4 AL 2, BM 3

hence area of trapezium is

= 1 15 4 3 4 2 4 302 2

sq. units

Also, 14tan A 2 A tan 22

Now, area of region enclosed by locus of

P is = 22 (10 5 5 2 5) (2) 30 = 70 4 5

50. (B,C,D)

Let 21 1(x ,y ) (at ,2at) .

Tangent at this point is 2yt x at .

Any point on this tangent is 2h ath,

t

.

Chord of contact with respect to the circle is 2

2h athx y at

[ 14 ] PT-V (ADV) RBA_11.11.2017


i.e. 2 y(aty a ) h x 0

t

Which is the family of the straight lines passing through the point of intersection of

yty a 0, and x 0t

.

So the fixed point is 2

a a,tt

2 2

axt

and 2ayt

51. (A,B,C,D)

(A)11 5 1 1sin sin sin cos sin Q12 12 12 12 2 6 4

(B)

9 4 1 16cosec sec cosec sec 4 Q10 5 10 5 sin18º cos36º ( 5 1)( 5 1)

(C) 4 4 1 1 3sin cos 1 sin² 1 Q8 8 2 4 4 4

(D) 22 42cos² 2cos ² 2cos ² 8(cos20º cos40º cos80º )9 9 9 =

1 Q8

52. (A, C, D)

[f (x) = 2 2

n

3Limx [cos x] n 1 n 3n 12

=3x2

+ x · [cosx] · 2 2

nLim n 1 n 3n 1

= 3x2

+ x[cosx] · 2 2

2 2n

n 1 (n 3n 1)Limn 1 n 3n 1

= 3x2

+ x[cosx] · 23

= 3x2 1 [cos x] [cos x]

=

23,

2x1

0x1

}0{2,

2x0 1

–1

O

3 2

2

y

x

y = cosx

PT-V (ADV) RBA_11.11.2017 [ 15 ]


f (x) =

3x x2 2 2

30 x2 2

Graph of f (x) in 3,2 2

3 2

3 4

– 2 –

–3 4

2–

y

xf(x)=0f(x)=

3x

2

O

From the graph it is clear that options (A), (C) and (D) are correct ]53. (7)

Centre of the given circle O(4,–3)

The circumcircle of PAB is (x 2)(x 4) (y 3)(y 3) 0

A

B

P(2,3)

O(4,-3)

2 2x y 6x 1 0 ........(i)

Director circle of ellipse is

2 2 2(x 5) (y 3) 9 b .......(ii)

From (i) and (ii) applying condition of orthogonality, we get

22[( 3)(5) 0( 3)] 1 25 b

2b 54 [b] 7

54 (4)Suppose px + qy =1 intersect the parabola y2 = 4ax at (at2,2at) then

21 2

2aq 2qpat q 2at 1 t tpa p

but we know that for co-normal points 2 2 21 1 2 2 3 3(at ,2at ), (at ,2at ),(at ,2at )

[ 16 ] PT-V (ADV) RBA_11.11.2017


1 2 3 32qt t t 0 tp

Hence 3rd point is

2

2

4aq 4aq,pp m 4

55. (2)

2 21 2 1 2a min (x x ) (y y )

where 1 1P(x ,y ) lies on 2y x 1 y x 1, y being positive

and 2 2Q(x ,y ) lies on 2y x 1

Here value of ‘a’ can be obtained if PQ is common normal to y2 = x – 1 and y = x2 + 1.Two parabolas are image of each other in the line y = xtherefore slope of common normal is – 1

Here, 21 1

1x 1 ( 1) x 5 / 44

and 11 1y ( 1)2 2

Hence P is

5 1,4 2 Q is

1 5,2 4

2 25 1 1 5 9 3PQ 24 2 2 4 16 2 2

2 9a PQ8

hence, 2 2a 1 2

56. (2)

In the triangle, tan A.tanB. tanC tanA tanB tanC

or

1 2k 1 4k 1 3. 3k2 2 2 2

or

28k 6k 1 3 6k

8 2 28k 6k 1 12 24K 28k 18k 11 0

1 11(2k 1)(4k 11) 0 k or k2 4 [k] = 2

PT-V (ADV) RBA_11.11.2017 [ 17 ]


57. (4)

Three vertices lies on the circle 2 2x y 25

Let orthocentre is H(h,k) then

h 5 sin 5cos 3 k 5 sin 5cos 4

Hence

h k 7 h k 1sin ,cos

10 10 2 2(h k 7) (h k 1) 100

2 2(x y 7) (x y 1) 100 –7, 1, 10 and | | 4

58. (6)

Take QPT

P Q

RT

SM

than RS r 1sinPS 3r 3

2 2cos

3

PT 2 2PQ 3

PT 2 212 3

PT 8 2 m = 8 , n = 2

59. (A - r); (B - s) ; (C - p) ; (D - q)

A. Eqn. of tangent at (2cos , 3 sin ) is x ycos sin 12 3

. Mid point (h,k) of portion of

tangent between coordinate axes is 3(sec , cosec )

2

3h sec , k cosec2

2

2 2 21 3 4 31 4h 2k h k

B. Eqn. of chord having mid point (x1, y1) is 2 2

1 1 1 1x x y y x y4 3 4 3

Eqn of pair of lines joining origin to extremities of the chord is

2

1 12 2

2 21 1

x x y yx y 4 31

x y4 34 3

The lines are mutually perpendicular

[ 18 ] PT-V (ADV) RBA_11.11.2017


2 2coeff. of x coeff of y 0

2 22 2 2 2 2 21 1 1 1 1 1x y x x y y1 1 0

4 4 3 16 3 4 3 9

22 2 2 21 1 1 1x y x y7 12

4 3 16 9

C. Eqn. of chord of contact from (h,k) is hx ky 14 3 which passes through the focus (1, 0)

or (–1, 0)

2h 1 h 164

D. Eqn. of normal at 22(2cos , 3 sin ) is 2x sec 3ycosec 2 3 1 ..... (1)

Eqn. of chord with mid point (x1, y1) is 2 2

1 1 1 1x x y y x y4 3 4 3

..... (2)

Comparing (1) and (2), 2 21 11 1

2sec 3 cosec 1x yx / 4 y / 34 3

2 2 2 21 1 1 1

1 1

x y x y8 3 3cos , sinx 4 3 y 4 3

22 21 1

2 21 1

x y64 27 1x y 4 3

60. (A - q); (B - r); (C - p); (D - s)

Equation of tangent to parabola = ay mxm

which passes through P(6,5) 1 1m or2 3

points of contact Q and R = (4,4) and (9,6) and area of 1PQR2

2 22C : (x 9) (y 6) (x 3y 9) 0 ; 2 2

1C : (x 4) (y 4) µ(x 2y 4) 0

now C2 & C1 pass through (1,0)2 2

2C : x y 28x 18y 27 0 ; 2 21C : x y 13x 2y 12 0

Centroid of PQR is 19,53

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PROGRESS TEST-5 · 2017-12-02 · [ 8 ] PT-V (ADV) RBA_11.11.2017 Mentors Eduserv: Parus Lok...

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